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1 year ago

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You drag a crate across a floor with a rope. The force applied is 750 N and the angle of the rope is 25.0° above the horizontalH

ow much work do you do on the crate when moving it 26.0 m?If you complete the job in 6.00 s, what power is developed?

1 answer:

frez [133]1 year ago

5 0

a.

The work done by a constant force along a rectilinear motion when the force and the displacement vector are not colinear is given by:

$W=F\cos\theta\cdot d$

where F is the magnitude of the force, theta is the angle between them and d is the distance.

The problen gives the following data:

The magnitude of the force 750 N.

The angle between the force and the displacement which is 25°

The distance, 26 m.

Plugging this in the formula we have:

$\begin{gathered} W=\left(750\right)\left(\cos25\right)\left(26\right) \\ W=17673 \end{gathered}$

Therefore the work done is 17673 J.

b)

The power is given by:

$P=\frac{W}{t}$

the problem states that the time it takes is 6 s. Then:

$\begin{gathered} P=\frac{17673}{6} \\ P=2945.5 \end{gathered}$

Therefore the power is 2945.5 W

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3 years ago

A beaker of water rests on an electronic balance that reads 875.0 g. A 2.75-cm-diameter solid copper ball attached to a string i

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a

The tension in the string is $T = 0.85 N$

b

The new balance reading is $M_b = 885.86 g$

Explanation:

From the question we are told that

The mass of the beaker of water is $m = 875 .0g$

The diameter of the copper ball is $d = 2.75 cm = \frac{2.75}{100} = 0.0275 m$

There are two forces acting on the copper ball

The first is the Buoyant force of the water pushing it up which is mathematically represented as

$F = \rho V g$

Where $\rho$ is the density of water which has value of $\rho = 1000 kg/m^3$

g is the acceleration due to gravity $g= 9.8 \ m/s^2$

$V$ is the volume of water displaced by the copper ball which is mathematically evaluated as

$V = \frac{4}{3} \pi r^3$

The radius r is $r = \frac{d}{3} = \frac{0.0275}{2} = 0.01375 m$

Substituting value

$V = \frac{4}{3} * 3.142 * (0.01375)^3$

$V = 1.08 9 *10^{-5 } m^3$

Substituting for F

$F = 1000 * 1.089 *10^{-5} * 9.8$

$F = 0.1067 N$

The second force is the weight of the copper ball which is mathematically represented as

$W_c = mg$

Now m is the mass which can be mathematically evaluated as

$m = \rho_c * V$

Where is the density of copper with value of $\rho_c = 8960 kg /m^3$

So $m = 8960 * 1.089 *10^{-5}$

$m = 0.0976$

So the weight of copper is

$W_c = 0.09756 * 9.8$

$W_c = 0.956 N$

Now the tension the string would be mathematically evaluated as

$T = W_c - F$

So $T = 0.956 - 0.1067$

$T = 0.85 N$

From this value we that the string is holding only 0.85 N of the copper weight thus (0.956 - 0.85 = 0.1065 N ) is being held by the balance

Now the mass equivalent of this weight is mathematically evaluated as

$m_z = \frac{1.0645 }{9.8 }$

$m_z = 0.01086 kg$

Converting to grams

$m_z = 10.86 g$

So the new balance reading is

$M_b = 875.0 +10.86$

$M_b = 885.86 g$

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